2025 New 1z0-830 Exam Questions Real Oracle Dumps
Course 2025 1z0-830 Test Prep Training Practice Exam Download
NEW QUESTION # 12
Given:
java
var hauteCouture = new String[]{ "Chanel", "Dior", "Louis Vuitton" };
var i = 0;
do {
System.out.print(hauteCouture[i] + " ");
} while (i++ > 0);
What is printed?
- A. An ArrayIndexOutOfBoundsException is thrown at runtime.
- B. Compilation fails.
- C. Chanel
- D. Chanel Dior Louis Vuitton
Answer: C
Explanation:
* Understanding the do-while Loop
* The do-while loopexecutes at least oncebefore checking the condition.
* The condition i++ > 0 increments iafterchecking.
* Step-by-Step Execution
* Iteration 1:
* i = 0
* Prints: "Chanel"
* i++ updates i to 1
* Condition 1 > 0is true, so the loop exits.
* Why Doesn't the Loop Continue?
* Since i starts at 0, the conditioni++ > 0 is false after the first iteration.
* The loopexits immediately after printing "Chanel".
* Final Output
nginx
Chanel
Thus, the correct answer is:Chanel
References:
* Java SE 21 - do-while Loop
* Java SE 21 - Post-Increment Behavior
NEW QUESTION # 13
Given:
java
DoubleStream doubleStream = DoubleStream.of(3.3, 4, 5.25, 6.66);
Predicate<Double> doublePredicate = d -> d < 5;
System.out.println(doubleStream.anyMatch(doublePredicate));
What is printed?
- A. Compilation fails
- B. An exception is thrown at runtime
- C. false
- D. true
- E. 3.3
Answer: A
Explanation:
In this code, there is a type mismatch between the DoubleStream and the Predicate<Double>.
* DoubleStream: A sequence of primitive double values.
* Predicate<Double>: A functional interface that operates on objects of type Double (the wrapper class), not on primitive double values.
The DoubleStream class provides a method anyMatch(DoublePredicate predicate), where DoublePredicate is a functional interface that operates on primitive double values. However, in the code, a Predicate<Double> is used instead of a DoublePredicate. This mismatch leads to a compilation error because anyMatch cannot accept a Predicate<Double> when working with a DoubleStream.
To correct this, the predicate should be defined as a DoublePredicate to match the primitive double type:
java
DoubleStream doubleStream = DoubleStream.of(3.3, 4, 5.25, 6.66);
DoublePredicate doublePredicate = d -> d < 5;
System.out.println(doubleStream.anyMatch(doublePredicate));
With this correction, the code will compile and print true because there are elements in the stream (e.g., 3.3 and 4.0) that are less than 5.
NEW QUESTION # 14
Given:
java
public class OuterClass {
String outerField = "Outer field";
class InnerClass {
void accessMembers() {
System.out.println(outerField);
}
}
public static void main(String[] args) {
System.out.println("Inner class:");
System.out.println("------------");
OuterClass outerObject = new OuterClass();
InnerClass innerObject = new InnerClass(); // n1
innerObject.accessMembers(); // n2
}
}
What is printed?
- A. Compilation fails at line n1.
- B. Nothing
- C. An exception is thrown at runtime.
- D. markdown
Inner class:
------------
Outer field - E. Compilation fails at line n2.
Answer: A
Explanation:
* Understanding Inner Classes in Java
* Aninner class (non-static nested class)requires an instance of the outer classbefore it can be instantiated.
* Incorrect instantiationof the inner class at n1:
java
InnerClass innerObject = new InnerClass(); // Compilation error
* Since InnerClass is anon-staticinner class, itmust be created from an instance of OuterClass.
* Correct Way to Instantiate the Inner Class
java
OuterClass outerObject = new OuterClass();
OuterClass.InnerClass innerObject = outerObject.new InnerClass(); // Correct
* Thiscorrectly associatesthe inner class with an instance of OuterClass.
* Why Does Compilation Fail?
* The error occurs atline n1because InnerClass is beinginstantiated incorrectly.
Thus, the correct answer is:Compilation fails at line n1.
References:
* Java SE 21 - Nested and Inner Classes
* Java SE 21 - Accessing Outer Class Members
NEW QUESTION # 15
Given:
java
Stream<String> strings = Stream.of("United", "States");
BinaryOperator<String> operator = (s1, s2) -> s1.concat(s2.toUpperCase()); String result = strings.reduce("-", operator); System.out.println(result); What is the output of this code fragment?
- A. United-STATES
- B. -UNITEDSTATES
- C. UNITED-STATES
- D. UnitedStates
- E. United-States
- F. -UnitedStates
- G. -UnitedSTATES
Answer: G
Explanation:
In this code, a Stream of String elements is created containing "United" and "States". A BinaryOperator<String> named operator is defined to concatenate the first string (s1) with the uppercase version of the second string (s2). The reduce method is then used with "-" as the identity value and operator as the accumulator.
The reduce method processes the elements of the stream as follows:
* Initial Identity Value: "-"
* First Iteration:
* Accumulator Operation: "-".concat("United".toUpperCase())
* Result: "-UNITED"
* Second Iteration:
* Accumulator Operation: "-UNITED".concat("States".toUpperCase())
* Result: "-UNITEDSTATES"
Therefore, the final result stored in result is "-UNITEDSTATES", and the output of theSystem.out.println (result); statement is -UNITEDSTATES.
NEW QUESTION # 16
Given:
java
StringBuffer us = new StringBuffer("US");
StringBuffer uk = new StringBuffer("UK");
Stream<StringBuffer> stream = Stream.of(us, uk);
String output = stream.collect(Collectors.joining("-", "=", ""));
System.out.println(output);
What is the given code fragment's output?
- A. US=UK
- B. An exception is thrown.
- C. Compilation fails.
- D. =US-UK
- E. US-UK
- F. -US=UK
Answer: D
Explanation:
In this code, two StringBuffer objects, us and uk, are created with the values "US" and "UK", respectively. A stream is then created from these objects using Stream.of(us, uk).
The collect method is used with Collectors.joining("-", "=", ""). The joining collector concatenates the elements of the stream into a single String with the following parameters:
* Delimiter ("-"):Inserted between each element.
* Prefix ("="):Inserted at the beginning of the result.
* Suffix (""):Inserted at the end of the result.
Therefore, the elements "US" and "UK" are concatenated with "-" between them, resulting in "US-UK". The prefix "=" is added at the beginning, resulting in the final output =US-UK.
NEW QUESTION # 17
Given:
java
DoubleSummaryStatistics stats1 = new DoubleSummaryStatistics();
stats1.accept(4.5);
stats1.accept(6.0);
DoubleSummaryStatistics stats2 = new DoubleSummaryStatistics();
stats2.accept(3.0);
stats2.accept(8.5);
stats1.combine(stats2);
System.out.println("Sum: " + stats1.getSum() + ", Max: " + stats1.getMax() + ", Avg: " + stats1.getAverage()); What is printed?
- A. Sum: 22.0, Max: 8.5, Avg: 5.0
- B. An exception is thrown at runtime.
- C. Sum: 22.0, Max: 8.5, Avg: 5.5
- D. Compilation fails.
Answer: C
Explanation:
The DoubleSummaryStatistics class in Java is part of the java.util package and is used to collect and summarize statistics for a stream of double values. Let's analyze how the methods work:
* Initialization and Data Insertion
* stats1.accept(4.5); # Adds 4.5 to stats1.
* stats1.accept(6.0); # Adds 6.0 to stats1.
* stats2.accept(3.0); # Adds 3.0 to stats2.
* stats2.accept(8.5); # Adds 8.5 to stats2.
* Combining stats1 and stats2
* stats1.combine(stats2); merges stats2 into stats1, resulting in one statistics summary containing all values {4.5, 6.0, 3.0, 8.5}.
* Calculating Output Values
* Sum= 4.5 + 6.0 + 3.0 + 8.5 = 22.0
* Max= 8.5
* Average= (22.0) / 4 = 5.5
Thus, the output is:
yaml
Sum: 22.0, Max: 8.5, Avg: 5.5
References:
* Java SE 21 & JDK 21 - DoubleSummaryStatistics
* Java SE 21 - Streams and Statistical Operations
NEW QUESTION # 18
What is the output of the following snippet? (Assume the file exists)
java
Path path = Paths.get("C:\\home\\joe\\foo");
System.out.println(path.getName(0));
- A. Compilation error
- B. home
- C. C
- D. C:
- E. IllegalArgumentException
Answer: B
Explanation:
In Java's java.nio.file package, the Path class represents a file path in a file system. The Paths.get(String first, String... more) method is used to create a Path instance by converting a path string or URI.
In the provided code snippet, the Path object path is created with the string "C:\\home\\joe\\foo". This represents an absolute path on a Windows system.
The getName(int index) method of the Path class returns a name element of the path as a Path object. The index is zero-based, where index 0 corresponds to the first element in the path's name sequence. It's important to note that the root component (e.g., "C:\" on Windows) is not considered a name element and is not included in this sequence.
Therefore, for the path "C:\\home\\joe\\foo":
* Root Component:"C:\"
* Name Elements:
* Index 0: "home"
* Index 1: "joe"
* Index 2: "foo"
When path.getName(0) is called, it returns the first name element, which is "home". Thus, the output of the System.out.println statement is home.
NEW QUESTION # 19
Given:
java
public class ExceptionPropagation {
public static void main(String[] args) {
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
}
static int thrower() {
try {
int i = 0;
return i / i;
} catch (NumberFormatException e) {
System.out.print("Rose");
return -1;
} finally {
System.out.print("Beaujolais Nouveau, ");
}
}
}
What is printed?
- A. Rose
- B. Beaujolais Nouveau, Chablis, Saint-Emilion
- C. Saint-Emilion
- D. Beaujolais Nouveau, Chablis, Dom Perignon, Saint-Emilion
Answer: B
Explanation:
* Analyzing the thrower() Method Execution
java
int i = 0;
return i / i;
* i / i evaluates to 0 / 0, whichthrows ArithmeticException (/ by zero).
* Since catch (NumberFormatException e) doesnot matchArithmeticException, it is skipped.
* The finally block always executes, printing:
nginx
Beaujolais Nouveau,
* The exceptionpropagates backto main().
* Handling the Exception in main()
java
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
* Since thrower() throws ArithmeticException, it is caught by catch (Exception e).
* "Chablis, "is printed.
* Thefinally block always executes, printing "Saint-Emilion".
* Final Output
nginx
Beaujolais Nouveau, Chablis, Saint-Emilion
Thus, the correct answer is:Beaujolais Nouveau, Chablis, Saint-Emilion
References:
* Java SE 21 - Exception Handling
* Java SE 21 - finally Block Execution
NEW QUESTION # 20
You are working on a module named perfumery.shop that depends on another module named perfumery.
provider.
The perfumery.shop module should also make its package perfumery.shop.eaudeparfum available to other modules.
Which of the following is the correct file to declare the perfumery.shop module?
- A. File name: module-info.java
java
module perfumery.shop {
requires perfumery.provider;
exports perfumery.shop.eaudeparfum;
} - B. File name: module-info.perfumery.shop.java
java
module perfumery.shop {
requires perfumery.provider;
exports perfumery.shop.eaudeparfum.*;
} - C. File name: module.java
java
module shop.perfumery {
requires perfumery.provider;
exports perfumery.shop.eaudeparfum;
}
Answer: A
Explanation:
* Correct module descriptor file name
* A module declaration must be placed inside a file namedmodule-info.java.
* The incorrect filename module-info.perfumery.shop.javais invalid(Option A).
* The incorrect filename module.javais invalid(Option C).
* Correct module declaration
* The module declaration must match the name of the module (perfumery.shop).
* The requires perfumery.provider; directive specifies that perfumery.shop depends on perfumery.
provider.
* The exports perfumery.shop.eaudeparfum; statement allows the perfumery.shop.eaudeparfum package to beaccessible by other modules.
* The incorrect syntax exports perfumery.shop.eaudeparfum.*; in Option A isinvalid, as wildcards (*) arenot allowedin module exports.
Thus, the correct answer is:File name: module-info.java
References:
* Java SE 21 - Modules
* Java SE 21 - module-info.java File
NEW QUESTION # 21
Given:
java
StringBuilder result = Stream.of("a", "b")
.collect(
() -> new StringBuilder("c"),
StringBuilder::append,
(a, b) -> b.append(a)
);
System.out.println(result);
What is the output of the given code fragment?
- A. bca
- B. cbca
- C. cacb
- D. cba
- E. acb
- F. bac
- G. abc
Answer: D
Explanation:
In this code, a Stream containing the elements "a" and "b" is processed using the collect method. The collect method is a terminal operation that performs a mutable reduction on the elements of the stream using a Collector. In this case, custom implementations for the supplier, accumulator, and combiner are provided.
Components of the collect Method:
* Supplier:
* () -> new StringBuilder("c")
* This supplier creates a new StringBuilder initialized with the string "c".
* Accumulator:
* StringBuilder::append
* This accumulator appends each element of the stream to the StringBuilder.
* Combiner:
* (a, b) -> b.append(a)
* This combiner is used in parallel stream operations to merge two StringBuilder instances. It appends the contents of a to b.
Execution Flow:
* Stream Elements:"a", "b"
* Initial StringBuilder:"c"
* Accumulation:
* The first element "a" is appended to "c", resulting in "ca".
* The second element "b" is appended to "ca", resulting in "cab".
* Combiner:
* In this sequential stream, the combiner is not utilized. The combiner is primarily used in parallel streams to merge partial results.
Final Result:
The StringBuilder contains "cab". Therefore, the output of the program is:
nginx
cab
NEW QUESTION # 22
Given:
java
Runnable task1 = () -> System.out.println("Executing Task-1");
Callable<String> task2 = () -> {
System.out.println("Executing Task-2");
return "Task-2 Finish.";
};
ExecutorService execService = Executors.newCachedThreadPool();
// INSERT CODE HERE
execService.awaitTermination(3, TimeUnit.SECONDS);
execService.shutdownNow();
Which of the following statements, inserted in the code above, printsboth:
"Executing Task-2" and "Executing Task-1"?
- A. execService.submit(task2);
- B. execService.execute(task2);
- C. execService.execute(task1);
- D. execService.submit(task1);
- E. execService.call(task2);
- F. execService.run(task2);
- G. execService.run(task1);
- H. execService.call(task1);
Answer: A,D
Explanation:
* Understanding ExecutorService Methods
* execute(Runnable command)
* Runs the task but only supports Runnable (not Callable).
* #execService.execute(task2); fails because task2 is Callable<String>.
* submit(Runnable task)
* Submits a Runnable task for execution.
* execService.submit(task1); executes "Executing Task-1".
* submit(Callable<T> task)
* Submits a Callable<T> task for execution.
* execService.submit(task2); executes "Executing Task-2".
* call() Does Not Exist in ExecutorService
* #execService.call(task1); and execService.call(task2); are invalid.
* run() Does Not Exist in ExecutorService
* #execService.run(task1); and execService.run(task2); are invalid.
* Correct Code to Print Both Messages:
java
execService.submit(task1);
execService.submit(task2);
Thus, the correct answer is:execService.submit(task1); execService.submit(task2); References:
* Java SE 21 - ExecutorService
* Java SE 21 - Callable and Runnable
NEW QUESTION # 23
Given:
java
double amount = 42_000.00;
NumberFormat format = NumberFormat.getCompactNumberInstance(Locale.FRANCE, NumberFormat.Style.
SHORT);
System.out.println(format.format(amount));
What is the output?
- A. 42 000,00 €
- B. 42 k
- C. 0
- D. 42000E
Answer: B
Explanation:
In this code, a double variable amount is initialized to 42,000.00. The NumberFormat.
getCompactNumberInstance(Locale.FRANCE, NumberFormat.Style.SHORT) method is used to obtain a compact number formatter for the French locale with the short style. The format method is then called to format the amount.
The compact number formatting is designed to represent numbers in a shorter form, based on the patterns provided for a given locale. In the French locale, the short style represents thousands with a lowercase 'k'.
Therefore, 42,000 is formatted as 42 k.
* Option Evaluations:
* A. 42000E: This format is not standard in the French locale for compact number formatting.
* B. 42 000,00 €: This represents the number as a currency with two decimal places, which is not the compact form.
* C. 42000: This is the plain number without any formatting, which does not match the compact number format.
* D. 42 k: This is the correct compact representation of 42,000 in the French locale with the short style.
Thus, option D (42 k) is the correct output.
NEW QUESTION # 24
Given:
java
public class BoomBoom implements AutoCloseable {
public static void main(String[] args) {
try (BoomBoom boomBoom = new BoomBoom()) {
System.out.print("bim ");
throw new Exception();
} catch (Exception e) {
System.out.print("boom ");
}
}
@Override
public void close() throws Exception {
System.out.print("bam ");
throw new RuntimeException();
}
}
What is printed?
- A. bim bam followed by an exception
- B. Compilation fails.
- C. bim boom
- D. bim bam boom
- E. bim boom bam
Answer: D
Explanation:
* Understanding Try-With-Resources (AutoCloseable)
* BoomBoom implements AutoCloseable, meaning its close() method isautomatically calledat the end of the try block.
* Step-by-Step Execution
* Step 1: Enter Try Block
java
try (BoomBoom boomBoom = new BoomBoom()) {
System.out.print("bim ");
throw new Exception();
}
* "bim " is printed.
* Anexception (Exception) is thrown, butbefore it is handled, the close() method is executed.
* Step 2: close() is Called
java
@Override
public void close() throws Exception {
System.out.print("bam ");
throw new RuntimeException();
}
* "bam " is printed.
* A new RuntimeException is thrown, but it doesnot override the existing Exception yet.
* Step 3: Exception Handling
java
} catch (Exception e) {
System.out.print("boom ");
}
* The catch (Exception e)catches the original Exception from the try block.
* "boom " is printed.
* Final Output
nginx
bim bam boom
* Theoriginal Exception is caught, not the RuntimeException from close().
* TheRuntimeException from close() is ignoredbecause thecatch block is already handling Exception.
Thus, the correct answer is:bim bam boom
References:
* Java SE 21 - Try-With-Resources
* Java SE 21 - AutoCloseable Interface
NEW QUESTION # 25
Which three of the following are correct about the Java module system?
- A. The unnamed module can only access packages defined in the unnamed module.
- B. If a request is made to load a type whose package is not defined in any known module, then the module system will attempt to load it from the classpath.
- C. If a package is defined in both a named module and the unnamed module, then the package in the unnamed module is ignored.
- D. Code in an explicitly named module can access types in the unnamed module.
- E. The unnamed module exports all of its packages.
- F. We must add a module descriptor to make an application developed using a Java version prior to SE9 run on Java 11.
Answer: B,C,E
Explanation:
The Java Platform Module System (JPMS), introduced in Java 9, modularizes the Java platform and applications. Understanding the behavior of named and unnamed modules is crucial.
* B. The unnamed module exports all of its packages.
Correct. The unnamed module, which includes all code on the classpath, exports all of its packages. This means that any code can access the public types in these packages. However, the unnamed module cannot be explicitly required by named modules.
* C. If a package is defined in both a named module and the unnamed module, then the package in the unnamed module is ignored.
Correct. In cases where a package is present in both a named module and the unnamed module, the version in the named module takes precedence. The package in the unnamed module is ignored to maintain module integrity and avoid conflicts.
* F. If a request is made to load a type whose package is not defined in any known module, then the module system will attempt to load it from the classpath.
Correct. When the module system cannot find a requested type in any known module, it defaults to searching the classpath (i.e., the unnamed module) to locate the type.
Incorrect Options:
* A. Code in an explicitly named module can access types in the unnamed module.
Incorrect. Named modules cannot access types in the unnamed module. The unnamed module can read from named modules, but the reverse is not allowed to ensure strong encapsulation.
* D. We must add a module descriptor to make an application developed using a Java version prior to SE9 run on Java 11.
Incorrect. Adding a module descriptor (module-info.java) is not mandatory for applications developed before Java 9 to run on Java 11. Such applications can run in the unnamed module without modification.
* E. The unnamed module can only access packages defined in the unnamed module.
Incorrect. The unnamed module can access all packages exported by all named modules, in addition to its own packages.
NEW QUESTION # 26
Given:
java
String colors = "red\n" +
"green\n" +
"blue\n";
Which text block can replace the above code?
- A. java
String colors = """
red
green
blue
"""; - B. java
String colors = """
red \s
green\s
blue \s
"""; - C. None of the propositions
- D. java
String colors = """
red \
green\
blue \
"""; - E. java
String colors = """
red \t
green\t
blue \t
""";
Answer: A
Explanation:
* Understanding Multi-line Strings in Java (""" Text Blocks)
* Java 13 introducedtext blocks ("""), allowing multi-line stringswithout needing explicit \n for new lines.
* In a text block,each line is preserved as it appears in the source code.
* Analyzing the Options
* Option A: \ (Backslash Continuation)
* The backslash (\) at the end of a lineprevents a new line from being added, meaning:
nginx
red green blue
* Incorrect.
* Option B: \s (Whitespace Escape)
* \s represents asingle space,not a new line.
* The output would be:
nginx
red green blue
* Incorrect.
* Option C: \t (Tab Escape)
* \t inserts atab, not a new line.
* The output would be:
nginx
red green blue
* Incorrect.
* Option D: Correct Text Block
java
String colors = """
red
green
blue
""";
* Thispreserves the new lines, producing:
nginx
red
green
blue
* Correct.
Thus, the correct answer is:"String colors = """ red green blue """."
References:
* Java SE 21 - Text Blocks
* Java SE 21 - String Formatting
NEW QUESTION # 27
Given:
java
LocalDate localDate = LocalDate.of(2020, 8, 8);
Date date = java.sql.Date.valueOf(localDate);
DateFormat formatter = new SimpleDateFormat(/* pattern */);
String output = formatter.format(date);
System.out.println(output);
It's known that the given code prints out "August 08".
Which of the following should be inserted as the pattern?
- A. MM dd
- B. MMMM dd
- C. MM d
- D. MMM dd
Answer: B
Explanation:
To achieve the output "August 08", the SimpleDateFormat pattern must format the month in its full textual form and the day as a two-digit number.
* Pattern Analysis:
* MMMM: Represents the full name of the month (e.g., "August").
* dd: Represents the day of the month as a two-digit number, with leading zeros if necessary (e.g.,
"08").
Therefore, the correct pattern to produce the desired output is MMMM dd.
* Option Evaluations:
* A. MM d: Formats the month as a two-digit number and the day as a single or two-digit number without leading zeros. For example, "08 8".
* B. MM dd: Formats the month and day both as two-digit numbers. For example, "08 08".
* C. MMMM dd: Formats the month as its full name and the day as a two-digit number. For example, "August 08".
* D. MMM dd: Formats the month as its abbreviated name and the day as a two-digit number. For example, "Aug 08".
Thus, option C (MMMM dd) is the correct choice to match the output "August 08".
NEW QUESTION # 28
Which two of the following aren't the correct ways to create a Stream?
- A. Stream stream = Stream.of();
- B. Stream stream = Stream.ofNullable("a");
- C. Stream stream = Stream.empty();
- D. Stream stream = Stream.generate(() -> "a");
- E. Stream stream = new Stream();
- F. Stream<String> stream = Stream.builder().add("a").build();
Answer: E,F
NEW QUESTION # 29
Given:
java
String s = " ";
System.out.print("[" + s.strip());
s = " hello ";
System.out.print("," + s.strip());
s = "h i ";
System.out.print("," + s.strip() + "]");
What is printed?
- A. [ , hello ,hi ]
- B. [,hello,h i]
- C. [,hello,hi]
- D. [ ,hello,h i]
Answer: B
Explanation:
In this code, the strip() method is used to remove leading and trailing whitespace from strings. The strip() method, introduced in Java 11, is Unicode-aware and removes all leading and trailing characters that are considered whitespace according to the Unicode standard.
docs.oracle.com
Analysis of Each Statement:
* First Statement:
java
String s = " ";
System.out.print("[" + s.strip());
* The string s contains four spaces.
* Applying s.strip() removes all leading and trailing spaces, resulting in an empty string.
* The output is "[" followed by the empty string, so the printed result is "[".
* Second Statement:
java
s = " hello ";
System.out.print("," + s.strip());
* The string s is now " hello ".
* Applying s.strip() removes all leading and trailing spaces, resulting in "hello".
* The output is "," followed by "hello", so the printed result is ",hello".
* Third Statement:
java
s = "h i ";
System.out.print("," + s.strip() + "]");
* The string s is now "h i ".
* Applying s.strip() removes the trailing spaces, resulting in "h i".
* The output is "," followed by "h i" and then "]", so the printed result is ",h i]".
Combined Output:
Combining all parts, the final output is:
css
[,hello,h i]
NEW QUESTION # 30
Given:
java
interface A {
default void ma() {
}
}
interface B extends A {
static void mb() {
}
}
interface C extends B {
void ma();
void mc();
}
interface D extends C {
void md();
}
interface E extends D {
default void ma() {
}
default void mb() {
}
default void mc() {
}
}
Which interface can be the target of a lambda expression?
- A. B
- B. D
- C. C
- D. A
- E. E
- F. None of the above
Answer: F
Explanation:
In Java, a lambda expression can be used where a target type is a functional interface. A functional interface is an interface that contains exactly one abstract method. This concept is also known as a Single Abstract Method (SAM) type.
Analyzing each interface:
* Interface A: Contains a single default method ma(). Since default methods are not abstract, A has no abstract methods.
* Interface B: Extends A and adds a static method mb(). Static methods are also not abstract, so B has no abstract methods.
* Interface C: Extends B and declares two abstract methods: ma() (which overrides the default method from A) and mc(). Therefore, C has two abstract methods.
* Interface D: Extends C and adds another abstract method md(). Thus, D has three abstract methods.
* Interface E: Extends D and provides default implementations for ma(), mb(), and mc(). However, it does not provide an implementation for md(), leaving it as the only abstract method in E.
For an interface to be a functional interface, it must have exactly one abstract method. In this case, E has one abstract method (md()), so it qualifies as a functional interface. However, the question asks which interface can be the target of a lambda expression. Since E is a functional interface, it can be the target of a lambda expression.
Therefore, the correct answer is D (E).
NEW QUESTION # 31
Given:
java
Map<String, Integer> map = Map.of("b", 1, "a", 3, "c", 2);
TreeMap<String, Integer> treeMap = new TreeMap<>(map);
System.out.println(treeMap);
What is the output of the given code fragment?
- A. {b=1, c=2, a=3}
- B. {a=1, b=2, c=3}
- C. {c=1, b=2, a=3}
- D. {b=1, a=3, c=2}
- E. {c=2, a=3, b=1}
- F. {a=3, b=1, c=2}
- G. Compilation fails
Answer: F
Explanation:
In this code, a Map named map is created using Map.of with the following key-value pairs:
* "b": 1
* "a": 3
* "c": 2
The Map.of method returns an immutable map containing these mappings.
Next, a TreeMap named treeMap is instantiated by passing the map to its constructor:
java
TreeMap<String, Integer> treeMap = new TreeMap<>(map);
The TreeMap constructor with a Map parameter creates a new tree map containing the same mappings as the given map, ordered according to the natural ordering of its keys. In Java, the natural ordering for String keys is lexicographical order.
Therefore, the TreeMap will store the entries in the following order:
* "a": 3
* "b": 1
* "c": 2
When System.out.println(treeMap); is executed, it outputs the TreeMap in its natural order, resulting in:
r
{a=3, b=1, c=2}
Thus, the correct answer is option F: {a=3, b=1, c=2}.
NEW QUESTION # 32
Given:
java
interface Calculable {
long calculate(int i);
}
public class Test {
public static void main(String[] args) {
Calculable c1 = i -> i + 1; // Line 1
Calculable c2 = i -> Long.valueOf(i); // Line 2
Calculable c3 = i -> { throw new ArithmeticException(); }; // Line 3
}
}
Which lines fail to compile?
- A. Line 2 only
- B. Line 3 only
- C. Line 1 and line 2
- D. Line 1 and line 3
- E. Line 2 and line 3
- F. The program successfully compiles
- G. Line 1 only
Answer: F
Explanation:
In this code, the Calculable interface defines a single abstract method calculate that takes an int parameter and returns a long. The main method contains three lambda expressions assigned to variables c1, c2, and c3 of type Calculable.
* Line 1:Calculable c1 = i -> i + 1;
This lambda expression takes an integer i and returns the result of i + 1. Since the expression i + 1 results in an int, and Java allows implicit widening conversion from int to long, this line compiles successfully.
* Line 2:Calculable c2 = i -> Long.valueOf(i);
Here, the lambda expression takes an integer i and returns the result of Long.valueOf(i). The Long.valueOf (int i) method returns a Long object. However, Java allows unboxing of the Long object to a long primitive type when necessary. Therefore, this line compiles successfully.
* Line 3:Calculable c3 = i -> { throw new ArithmeticException(); };
This lambda expression takes an integer i and throws an ArithmeticException. Since the method calculate has a return type of long, and throwing an exception is a valid way to exit the method without returning a value, this line compiles successfully.
Since all three lines adhere to the method signature defined in the Calculable interface and there are no type mismatches or syntax errors, the program compiles successfully.
NEW QUESTION # 33
Given:
java
var lyrics = """
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
""";
for ( int i = 0, int j = 3; i < j; i++ ) {
System.out.println( lyrics.lines()
.toList()
.get( i ) );
}
What is printed?
- A. vbnet
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose - B. Nothing
- C. An exception is thrown at runtime.
- D. Compilation fails.
Answer: D
Explanation:
* Error in for Loop Initialization
* The initialization part of a for loopcannot declare multiple variables with different types in a single statement.
* Error:
java
for (int i = 0, int j = 3; i < j; i++) {
* Fix:Declare variables separately:
java
for (int i = 0, j = 3; i < j; i++) {
* lyrics.lines() in Java 21
* The lines() method of String returns aStream<String>, splitting the string by line breaks.
* Calling .toList() on a streamconverts it to a list.
* Valid Code After Fixing the Loop:
java
var lyrics = """
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
""";
for (int i = 0, j = 3; i < j; i++) {
System.out.println(lyrics.lines()
toList()
get(i));
}
* Expected Output After Fixing:
vbnet
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - String.lines()
* Java SE 21 - for Statement Rules
NEW QUESTION # 34
......
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